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受欢迎的微积分 >

integral of 2400xsqrt(3+2x)

解答

\int 2400 x 3 + 2 x d x

解答

600 (\frac{2}{5} (2 x + 3)^{\frac{5}{2}} - 2 (2 x + 3)^{\frac{3}{2}}) + C

求解步骤

\int 2400 x 3 + 2 x d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2400 \cdot \int x 3 + 2 x d x

使用换元积分法

= 2400 \cdot \int \frac{u ( u - 3 )}{4} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2400 \cdot \frac{1}{4} \cdot \int u (u - 3) d u

乘开

u (u - 3) : u^{\frac{3}{2}} - 3 u

= 2400 \cdot \frac{1}{4} \cdot \int u^{\frac{3}{2}} - 3 u d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2400 \cdot \frac{1}{4} (\int u^{\frac{3}{2}} d u - \int 3 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 3 u d u = 2 u^{\frac{3}{2}}

= 2400 \cdot \frac{1}{4} (\frac{2}{5} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}})

u = 3 + 2 x

代回

= 2400 \cdot \frac{1}{4} (\frac{2}{5} (3 + 2 x)^{\frac{5}{2}} - 2 (3 + 2 x)^{\frac{3}{2}})

化简

2400 \cdot \frac{1}{4} (\frac{2}{5} (3 + 2 x)^{\frac{5}{2}} - 2 (3 + 2 x)^{\frac{3}{2}}) : 600 (\frac{2}{5} (2 x + 3)^{\frac{5}{2}} - 2 (2 x + 3)^{\frac{3}{2}})

= 600 (\frac{2}{5} (2 x + 3)^{\frac{5}{2}} - 2 (2 x + 3)^{\frac{3}{2}})

解答补常数

= 600 (\frac{2}{5} (2 x + 3)^{\frac{5}{2}} - 2 (2 x + 3)^{\frac{3}{2}}) + C

作图

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